Mathematics for the interested outsider

Cauchy’s Invariant Rule

An immediate corollary of the chain rule is another piece of “syntactic sugar”.

If we have functions and for some open regions and so that the image is contained in , we can compose the two functions to get a new function . In terms of formulas, we can choose coordinates on and write out both the function and the component functions . We get a formula for by substituting for in the formula for and write .

The language there seems a little convoluted, so I’d like to give an example. We might define a function for all points in the plane . This is all well and good, but we might want to talk about the function in polar coordinates. To this end, we may define and . These are the component functions describing a transformation from the region to the region where . We can substitute for and for in our formula for to get a new function with formula

This much is straightforward. The thing is, now we want to take differentials. What Cauchy’s invariant rule tells us is that we can calculate the differential of by not only substituting for , but also substituting for in the formula for . That is, if then we have the equivalence

In our particular example, we can easily calculate the differential of using our first formula:

or using our second formula:

We want to call both of these simply . But can we do so unambiguously? Indeed, if then we find

and if then we find

We substitute these into our formula for to find

just the same as if we calculated directly from the formula in terms of and .

That is, we can substitute our formulæ for the coordinate functions before taking the differential in terms of , or we can take the differential in terms of and then substitute our formulæ for the coordinate functions and their differentials into the result. Either way, we end up in the same place, so we don’t have to worry about ending up with two (or more!) “different” differentials of .

So, how do we verify this using the chain rule? Just write out the differentials out using partial derivatives. For example, we know that

and so on. So, performing our substitutions we can find:

The important part here is the passage from products of two partial derivatives to single partial derivatives of . This works out because when we consider differentials as linear transformations, the matrix entries are the partial derivatives. The composition of the linear transformations and is given by the product of these matrices, and the entries of the resulting matrix must (by uniqueness) be the partial derivatives of the composite function.

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This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

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